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30 tháng 8 2023
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
I
3
TV
31 tháng 3 2023
a) Gọi x²=a
=> 3a² - a - 234=0
∆=b² - 4ac= (-1)²-4×3×(-234)=2809
√∆=53
∆>0 nên pt có 2 nghiệm phân biệt
a1=-b+√∆/2a = -(-1)+53/2×3 =9
a2=-b-√∆/2a = -(-1)-53/2×3 =-26/3
Thay x²=a=9 =>x=3,x=-3
x²=a=-26/3 (loại)
Vậy nghiệm của pt là x =3, x=-3
TV
31 tháng 3 2023
d) (x+4)(x+5)(x+7)(x+8)=4
<=> (x+4)(x+8)(x+5)(x+7)=4
<=> (x²+8x+4x+32)(x²+7x+5x+35)=4
<=> (x²+12x+32)(x²+12x+35)=4
Đặt t=x²+12x+32
=> t(t+3)=4
<=> t²+3t-4=0
(a=1,b=3,c=-4)
a+b+c=1+3+(-4)=0
=> t1=1 ; t2= c/a =-4/1=-4
Thay t=x²+12x+32=1
=> x²+12x+31=0
∆=b²-4ac= 12² -4×1×31= 20
√∆=2√5
∆>0 nên pt có 2 nghiệm phân biệt
x1=-b+√∆/2a= -12+2√5/2×1= -6+√5
x2=-b-√∆/2a = -12-2√5/2×1= -6-√5
Thay t=x²+12x+32=-4
=> x²+12x+36=0
∆=b²-4ac= 12²-4×1×36=0
∆=0 nên pt có nghiệm kép
x1=x2= -b/2a= -12/2×1 = -6
Vậy nghiệm của pt là S={-6+√5 ; -6-√5; -6}
4 tháng 1
a: \(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=14\\2x+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=5-2x=5-2\cdot2=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}-x+2y=2\\2x-y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x+4y=4\\2x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=3\\x-2y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=1\\x=-2+2y=-2+2\cdot1=0\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}2x-y=13\\y-5=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-y=13\\y=-7+5=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=y+13=-2+13=11\\y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{11}{2}\\y=-2\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}3x+y=8\\2x-3y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9x+3y=24\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=25\\3x+y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{25}{11}\\y=8-3x=8-3\cdot\dfrac{25}{11}=8-\dfrac{75}{11}=\dfrac{13}{11}\end{matrix}\right.\)
T
14 tháng 1 2019
a) \(\left\{{}\begin{matrix}3x-4y=-2\\2x+y=6\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}3x-4y=-2\\8x+4y=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\3x-4y=-2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
31 tháng 12 2022
a: =>3x-4y=-2 và 8x+4y=24
=>11x=22 và 2x+y=6
=>x=2 và y=6-2x=6-2*2=2
b: 2x-y=0 và 3x+y=4
=>5x=4 và y=2x
=>x=4/5 và y=8/5
c: x+3y=-2 và x-y=-1
=>4y=-1 và x=y-1
=>y=-1/4 và x=-1/4-1=-5/4
d: x+y=3 và 4x-3y=-2
=>4x+4y=12 và 4x-3y=-2
=>7y=14 và x+y=3
=>y=2 và x=1
26 tháng 11 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x< >\dfrac{3}{2}y\\x< >-\dfrac{y}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\\\dfrac{-5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{20}{2x-3y}+\dfrac{25}{3x+y}=-10\\-\dfrac{20}{2x-3y}+\dfrac{12}{3x+y}=84\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{37}{3x+y}=74\\-\dfrac{5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\-\dfrac{5}{2x-3y}+3:\dfrac{1}{2}=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\\dfrac{-5}{2x-3y}=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=\dfrac{3}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}11x=\dfrac{7}{6}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{66}\\3y=2x+\dfrac{1}{3}=\dfrac{7}{33}+\dfrac{1}{3}=\dfrac{6}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{2}{11}\end{matrix}\right.\)(nhận)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x< >y-2\\x< >-y+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{7}{x-y+2}-\dfrac{5}{x+y-1}=\dfrac{9}{2}\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{14}{x-y+2}-\dfrac{10}{x+y-1}=9\\\dfrac{15}{x-y+2}+\dfrac{10}{x+y-1}=20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{29}{x-y+2}=29\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y+2=1\\3+\dfrac{2}{x+y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\\dfrac{2}{x+y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y=-1\\x+y-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=2\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)(nhận)
c:
ĐKXĐ: \(\left\{{}\begin{matrix}y< >2x\\y< >-x\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{3}{2x-y}-\dfrac{3}{x+y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=3\\2x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=6\\2x-y=3\end{matrix}\right.\)
=>x=2 và y=2x-3=4-3=1(nhận)
d:ĐKXĐ: \(\left\{{}\begin{matrix}x< >-y+1\\x< >\dfrac{1}{2}y-\dfrac{3}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}+\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{19}{x+y-1}=\dfrac{19}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y-1=2\\\dfrac{15}{2}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\\dfrac{5}{2x-y+3}=7-\dfrac{15}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=3\\2x-y+3=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\2x-y=-13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=-10\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=3-x=3+\dfrac{10}{3}=\dfrac{19}{3}\end{matrix}\right.\left(nhận\right)\)
e:
ĐKXĐ: \(x\ne\pm2y\)
\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{6}{x-2y}+\dfrac{8}{x+2y}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{6}{x+2y}=5\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}+4:\dfrac{-6}{5}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}=-1+4\cdot\dfrac{5}{6}=-1+\dfrac{10}{3}=\dfrac{7}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{35}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3}{70}\\2y=x-\dfrac{9}{7}=-\dfrac{87}{70}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{70}\\y=-\dfrac{87}{140}\end{matrix}\right.\left(nhận\right)\)
\(\Leftrightarrow2^{3x+3}\cdot7^y=2^{6x}\cdot7^{2x}\cdot5^{x-1}\)
=>3x+3=6x và x-1=0 và 2x=y
=>-3x=-3 và x=1 và y=2x
=>x=1 và y=2